Question

Time constant of a first order unit step input system is defined as the time at which unit step response reaches

• A. \(63.2%\) of steady state value
• B. \(66.7%\) of steady state value
• C. \(36.8%\) of steady state value
• D. \(33.3%\) of steady state value

Hint:

For a rising exponential response: At \(t = 1\tau\), response reaches \(63.2%\). For a decaying exponential response: At \(t = 1\tau\), response drops down to \(36.8%\) of its initial values. Keeping this distinction clear avoids confusion during evaluations!

Answer 1

The Correct Option is A

Concept: The standard transfer function layout of a first-order system is described as: \[ G(s) = \frac{1}{1 + \tau s} \] where \(\tau\) is the time constant of the system. When a unit step input \(R(s) = \frac{1}{s}\) is applied, the output response in the time domain is given by the expression: \[ c(t) = 1 - e^{-t/\tau} \quad (\text{for } t \ge 0) \] The time constant \(\tau\) determines how rapidly the system reaches its final steady-state value.

Step 1: Evaluate the time response function at exactly \(t = \tau\). Substitute the time value equal to one time constant, i.e., \(t = \tau\), into the output equation: \[ c(\tau) = 1 - e^{-\tau/\tau} = 1 - e^{-1} \]

Step 2: Compute the numerical percentage equivalent. The standard numerical value of mathematical constant \(e\) is approximately equal to \(2.71828\). Thus: \[ e^{-1} = \frac{1}{2.71828} \approx 0.3678 \] Substitute this back into the expression for \(c(\tau)\): \[ c(\tau) = 1 - 0.3678 = 0.6322 \] Expressing this fraction as a percentage of the final steady-state value (which is \(1\)): \[ \text{Percentage} = 0.6322 \times 100% = 63.2% \] Thus, the time constant represents the time taken for the transient response to reach \(63.2%\) of its ultimate steady-state target, matching option (A).